Question

If the wavelength of series limit of lyman series for helium ion in x then what will be the wavelength of series

Answer 1

Answer:

=R[

n

1

2

1

−

n

2

2

1

]

For limiting wavelength of Lyman series

n_{1} = 1, n_{2} = \inftyn

1

=1,n

2

=∞

\dfrac {1}{\lambda_{L}} = R

λ

L

1

=R

For limiting wavelength of Balmer series

n_{1} = 2, n_{2} = \inftyn

1

=2,n

2

=∞

\dfrac {1}{\lambda_{B}} = R\left (\dfrac {1}{4}\right ) \Rightarrow \lambda_{B} = \dfrac {4}{R}

λ

B

1

=R(

4

1

)⇒λ

B

=

R

4

\therefore \lambda_{B} = 4\lambda_{L} = 4\times 912\overset {\circ}{A}∴λ

B

=4λ

L

=4×912

A

∘

.